A liquid drop of radius R is broken up into N small droplets. The work done is proportional to

The correct option is N¹/³.

Let us derive the expression for the work done in breaking up a liquid drop into smaller droplets step-by-step.

Let the radius of the original large liquid drop be $R$ and the surface tension of the liquid be $T$.

The volume of the original large drop is given by:
$V_{\text{initial}}=\frac{4}{3}\pi R^3$

This large drop is broken into $N$ small droplets, each of radius $r$. Since the total volume of liquid remains constant:
$\frac{4}{3}\pi R^3=N·\left(\frac{4}{3}\pi r^3\right)$

Simplifying this equation, we get the relation between the radii:
$R^3=N{r}^3$
$r=\frac{R}{N^{1/3}}$

Now, let us calculate the change in surface area.
The initial surface area of the single large drop is:
$A_{\text{initial}}=4\pi R^2$

The final surface area of $N$ small droplets is:
$A_{\text{final}}=N·\left(4\pi r^2\right)$

Substituting $r=R{N}^{-1/3}$ into the final surface area expression:
$A_{\text{final}}=N·4\pi {\left(\frac{R}{N^{1/3}}\right)}^2=N·4\pi \frac{R^2}{N^{2/3}}=N^{1/3}·4\pi R^2$

The increase in surface area $\mathrm{\Delta A}$ is:
$\mathrm{\Delta A}=A_{\text{final}}-A_{\text{initial}}=4\pi R^2(N^{1/3}-1)$

The work done $W$ in this process is equal to the surface tension multiplied by the increase in surface area:
$W=T·\mathrm{\Delta A}=4\pi R^{2}T(N^{1/3}-1)$

For a large number of droplets ($N\gg 1$), the term $N^{1/3}-1\approx N^{1/3}$. Therefore, the work done is proportional to:
$W\propto N^{1/3}$